Three capacitors $(2.0 \mu \mathrm{F}, 3.0 \mu \mathrm{F}, 6.0 \mu \mathrm{F})$ are connected in series to an ac voltage source with amplitude \(12.0 \mathrm{V}\) and frequency \(6.3 \mathrm{kHz}\). (a) What are the peak voltages across each capacitor? (b) What is the peak current that flows in the circuit?

To determine the equivalent capacitance of the capacitors connected in series, we can use the formula for series capacitance: \( \dfrac{1}{C_eq} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3}\) Where, \(C_1 = 2.0\;\mu F\) (The first capacitor) \(C_2 = 3.0\;\mu F\) (The second capacitor) \(C_3 = 6.0\;\mu F\) (The third capacitor) Solve for \(C_eq\): \(C_eq = \dfrac{1}{(\dfrac{1}{2.0 \times 10^{-6}} + \dfrac{1}{3.0 \times 10^{-6}} + \dfrac{1}{6.0 \times 10^{-6}})}\) \(C_eq = 1.0 \mu F\)

Now we need to find the angular frequency (\(\omega\)) of the AC voltage source. The relationship between the angular frequency and frequency is: \(\omega = 2\pi f\) Where, \(f = 6.3\;kHz\) (Frequency of the AC voltage source) \(\omega = 2\pi \times 6.3 \times 10^{3} = 39600\;rad/s\)

Next, calculate the impedance (Z) of the circuit using the equivalent capacitance and the angular frequency: \(Z = \dfrac{1}{\omega C_eq}\) \(Z = \dfrac{1}{39600 \times 1.0 \times 10^{-6}}\) \(Z = 25.3\;\Omega\)

Now we can determine the peak current (I) in the circuit using Ohm's law: \(I = \dfrac{V}{Z}\) Where, \(V = 12.0\;V\) (Amplitude of the AC voltage source) \(I = \dfrac{12}{25.3} = 0.475\;A\) The peak current that flows in the circuit is \(0.475\;A\).

Finally, we can calculate the peak voltage across each capacitor using the peak current and individual capacitor's impedance: \(V_{C_1} = I \times Z_{C_1}\) \(V_{C_2} = I \times Z_{C_2}\) \(V_{C_3} = I \times Z_{C_3}\) Where, \(Z_{C_1} = \dfrac{1}{\omega C_1}\) \(Z_{C_2} = \dfrac{1}{\omega C_2}\) \(Z_{C_3} = \dfrac{1}{\omega C_3}\) Solve for each peak voltage: \(V_{C_1} = 0.475 \times \dfrac{1}{39600 \times 2.0 \times 10^{-6}} = 6.0\;V\) \(V_{C_2} = 0.475 \times \dfrac{1}{39600 \times 3.0 \times 10^{-6}} = 4.0\;V\) \(V_{C_3} = 0.475 \times \dfrac{1}{39600 \times 6.0 \times 10^{-6}} = 2.0\;V\) The peak voltages across the capacitors are \(6.0\;V\), \(4.0\;V\), and \(2.0\;V\) for \(C_1\), \(C_2\), and \(C_3\), respectively.